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Normal and shear components of the stress vector

  1. Aug 25, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the normal and shear components ([itex]\vec{\sigma}_{n},\vec{\tau}[/itex]) of the stress vector on the surface ABCD.

    Given:
    State of stress at a point
    [itex]\sigma_{xx}=3[/itex]
    [itex]\sigma_{yy}=0[/itex]
    [itex]\sigma_{zz}=-1[/itex]
    [itex]\sigma_{xy}=-\sqrt{2}[/itex]
    [itex]\sigma_{xz}=1000[/itex]

    Unit normal:
    [itex]\widehat{n}=\frac{1}{\sqrt{2}}\widehat{i}+\frac{1}{\sqrt{2}}\widehat{j}+0\widehat{k}[/itex]


    2. Relevant equations
    [itex]\vec{\sigma}_{n}=\vec{t}\cdot\widehat{n}[/itex]
    [itex]\vec{t}=\vec{\sigma}_{n}+\vec{\tau}[/itex]



    3. The attempt at a solution

    First I take the stress matrix and multiply it by the unit normal to find the traction components

    [itex]\sigma = \left[\begin{matrix}
    3 & -\sqrt{2} & 1000 \\
    -\sqrt{2} & 0 & -\sqrt{2} \\
    1000 & -\sqrt{2} & -1 \\
    \end{matrix}
    \right][/itex]

    [itex]\left[\begin{matrix}
    3 & -\sqrt{2} & 1000 \\
    -\sqrt{2} & 0 & -\sqrt{2} \\
    1000 & -\sqrt{2} & -1 \\
    \end{matrix}
    \right]
    \left[\begin{matrix}
    \frac{1}{-\sqrt{2}} \\
    \frac{1}{-\sqrt{2}} \\
    0 \\
    \end{matrix}
    \right]
    = \left[\begin{matrix}
    t_{x} \\
    t_{y} \\
    t_{z} \\
    \end{matrix}
    \right][/itex]

    This results in
    [itex]\vec{t}=
    \left[\begin{matrix}
    \frac{3}{\sqrt{2}}-1 \\
    -1 \\
    \frac{1000}{\sqrt{2}}-1 \\
    \end{matrix}
    \right][/itex]

    Then I find the normal component

    [itex]\vec{\sigma}_{n}=\left[ \frac{3}{\sqrt{2}}-1,-1,\frac{1000}{\sqrt{2}}-1\right] \cdot
    \left[\begin{matrix}
    \frac{1}{-\sqrt{2}} \\
    \frac{1}{-\sqrt{2}} \\
    0 \\
    \end{matrix}
    \right][/itex]

    [itex]\vec{\sigma}_{n} = \frac{3}{2}-\sqrt{2}[/itex]

    Using the hint that was provided, I solve for [itex]\vec{\tau}[/itex]
    [itex]\vec{\tau}=\vec{t}-\vec{\sigma}_n[/itex]

    Is this procedure correct thus far? I've seen elsewhere that
    [itex]\tau=\vec{t}-(\vec{t}\cdot \vec{n})\vec{n}[/itex]

    I'm a bit confused with the notation of the supplied equation for [itex]\sigma[/itex] because if you take the dot product of a vector with a unit normal you should only get a scalar. The notation shows that it is a vector.
     
  2. jcsd
  3. Aug 25, 2013 #2
    In your notation, [itex]\vec{\sigma}[/itex] is the stress tensor. The stress vector on a surface perpendicular to the unit normal [itex]\vec{n}[/itex] is given by:
    [tex]\vec{t}=\vec{\sigma}\centerdot\vec{n}[/tex]
    Note that [itex]\vec{\sigma}[/itex] is a second order tensor, and, when you dot it with the unit normal, you produce a vector (first order tensor) [itex]\vec{t}[/itex]. The vector [itex]\vec{t}[/itex] is not pointing perpendicular to the surface. If you want the component perpendicular to the surface, you need to dot it with the unit normal.
    Scalar component of stress vector perpendicular to the surface = [itex]\vec{n}\centerdot\vec{t}
    [/itex]
    The normal vector component of the stress vector on the surface is equal to the scalar component times the unit normal [itex]\vec{\sigma_n}=(\vec{n}\centerdot\vec{t})\vec{n}[/itex]
    The tangential vector component of the stress vector on the surface is equal to the stress vector minus the normal vector component:[itex]\vec{\tau}=\vec{t}-(\vec{n}\centerdot\vec{t})\vec{n}[/itex]

    Hope this helps.

    Chet
     
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